【leetcode】983. Minimum Cost For Tickets
作者:互联网
题目如下:
In a country popular for train travel, you have planned some train travelling one year in advance. The days of the year that you will travel is given as an array
days
. Each day is an integer from1
to365
.Train tickets are sold in 3 different ways:
- a 1-day pass is sold for
costs[0]
dollars;- a 7-day pass is sold for
costs[1]
dollars;- a 30-day pass is sold for
costs[2]
dollars.The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.
Return the minimum number of dollars you need to travel every day in the given list of
days
.
Example 1:
Input: days = [1,4,6,7,8,20], costs = [2,7,15] Output: 11 Explanation: For example, here is one way to buy passes that lets you travel your travel plan: On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1. On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9. On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20. In total you spent $11 and covered all the days of your travel.Example 2:
Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15] Output: 17 Explanation: For example, here is one way to buy passes that lets you travel your travel plan: On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30. On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31. In total you spent $17 and covered all the days of your travel.
Note:
1 <= days.length <= 365
1 <= days[i] <= 365
days
is in strictly increasing order.costs.length == 3
1 <= costs[i] <= 1000
解题思路:毕竟本人动态规划没有掌握的游刃有余,一时间想不出递推表达式。那就简单粗暴吧,对于任意一个days[i]来说,都有三种买票的方法,买1天,7天和30天,借助DFS的思想依次计算每一种方法的最小值,理论上是有3^365次方种组合,但是计算过程中可以舍去明显不符合条件的组合,因此此方法也能通过。
代码如下:
class Solution(object): def mincostTickets(self, days, costs): """ :type days: List[int] :type costs: List[int] :rtype: int """ res = len(days) * costs[0] queue = [(0,0)] #(inx,cost_inx,total) dp = [366*costs[2]] * (len(days) + 1) while len(queue) > 0: #print len(queue) inx,total = queue.pop(0) if inx == len(days): res = min(res,total) continue elif total > res: continue if dp[inx+1] > total + costs[0]: queue.insert(0,(inx+1, total + costs[0])) dp[inx+1] = total + costs[0] import bisect next_inx = bisect.bisect_left(days,days[inx]+7) if dp[next_inx] > total + costs[1]: queue.insert(0,(next_inx, total + costs[1])) next_inx = bisect.bisect_left(days, days[inx] + 30) if dp[next_inx] > total + costs[2]: queue.insert(0,(next_inx, total + costs[2])) return res
标签:Tickets,inx,983,travel,days,costs,Minimum,total,day 来源: https://www.cnblogs.com/seyjs/p/10356000.html